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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>We only need one form of <span class="process-math">\(v(x)\)</span> as long as <span class="process-math">\(v(x) y_1\)</span> is a solution linearly independent of <span class="process-math">\(y_1\text{.}\)</span> Setting <span class="process-math">\(d_1=1\)</span> and <span class="process-math">\(d_2=0\text{,}\)</span> we have</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y_2=x y_1=x e^{-\frac{b}{2} x}.
\end{equation*}
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<p class="continuation">Further,</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
W=\left| \begin{array}{ll}
y_1 &amp; y_2\\
y_1^{\prime} &amp; y_2^{\prime}
\end{array}\right|=
\left| \begin{array}{lll}
e^{-\frac{b}{2}x} &amp; \quad &amp; x e^{-\frac{b}{2}x}\\
-\frac{b}{2} e^{-\frac{b}{2} x} &amp; \quad &amp; e^{-\frac{b}{2}x}-\frac{b x}{2} e^{-\frac{b}{2} x}
\end{array} \right|=e^{-bx} \neq 0.
\end{equation*}
</div>
<p class="continuation">This means <span class="process-math">\(y_1\)</span> and <span class="process-math">\(y_2\)</span> are linear independent. So the general solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y=C_1 y_1+C_2 y_2=C_1 e^{-\frac{b}{2}x}+C_2 x  e^{-\frac{b}{2}x}.
\end{equation*}
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<span class="incontext"><a href="sec3_5.html#p-104" class="internal">in-context</a></span>
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